Home
Class 12
PHYSICS
A circular coil of radius 'r' having N t...

A circular coil of radius 'r' having N turns carries a current ''I''. What is its magnetic moment ?

Text Solution

Verified by Experts

Magnetic moment (M)=N I A
`M=N I (pi r^(2))" "(because A = pi r^(2))`
`therefore" "M= pi N I r^(2)`
Promotional Banner

Topper's Solved these Questions

  • MOVING CHARGES AND MAGNETISHM

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise SHORT ANSWER QUESTIONS|9 Videos
  • MOVING CHARGES AND MAGNETISHM

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise LONG ANSWER QUESTIONS|5 Videos
  • MOVING CHARGES AND MAGNETISHM

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise TEXTUAL EXAMPLES|18 Videos
  • MAGNETISM AND MATTER

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise Additional Exercises|19 Videos
  • NUCLEI

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise DAM SURE (VSAQ)|1 Videos

Similar Questions

Explore conceptually related problems

A 100 turn closely wound circular coil of radius 10 cm carries a current a 3.2 A. What is the magnetic moment of this coil ?

For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by, B = (mu_0 IR^2 N)/(2 (x^2 + R^2)^(3//2)) (a) Show that this reduces to the familiar result for field at the centre of the coil. (b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by, B = 0.72 (mu_0 NI)/(R ) . approximately [Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by, B=(mu_(0) IR^(2)N)/(2(x^(2)+R^(2))^(3//2)) a) Show that this reduces to the familiar result for field at the centre of the coil.

A circular coil of radius r having number of turns n and carrying a current A produces magnetic induction at its centre of magnitude B. B can be doubled by

A circular coil of radius 2 R is carrying current 'i'. The ratio of magnetic fields at the centre of the coil and at a point at a distance 6 R from the the centre of the coil on the axis of the coil is,